Let $R$ be a commutative Noetherian ring, $\mathfrak{a}$ an ideal of $R$ and $M$ a finitely generated $R$ -module. Let $t$ be a non-negative integer. It is known that if the local cohomology module $\text{H}_{\mathfrak{a}}^{i}\,\left( M \right)$ is finitely generated for all $i\,<\,t$ , then $\text{Ho}{{\text{m}}_{R}}\,\left( R/\mathfrak{a},\,\text{H}_{\mathfrak{a}}^{t}\,\left( M \right) \right)$ is finitely generated. In this paper it is shown that if $\text{H}_{\mathfrak{a}}^{i}\,\left( M \right)$ is Artinian for all $i\,<\,t$ , then $\text{Ho}{{\text{m}}_{R}}\,\left( R/\mathfrak{a},\,\text{H}_{\mathfrak{a}}^{t}\,\left( M \right) \right)$ need not be Artinian, but it has a finitely generated submodule $N$ such that $\text{Ho}{{\text{m}}_{R}}\left( R/\mathfrak{a},\text{H}_{\mathfrak{a}}^{t}\left( M \right) \right)/N$ is Artinian.

A non-zero module M having a minimal generator set contains a maximal submodule. If M is Artinian and all submodules of M have minimal generator sets then M is Noetherian; it follows that every left Artinian module of a left perfect ring is Noetherian. Every right Noetherian module of a left perfect ring is Artinian. It follows that a module over a left and right perfect ring (in particular, commutative) is Artinian if and only if it is Noetherian. We prove that a local ring is left perfect if and only if each left module has a minimal generator set.